Pretty standard Project Euler solution. You only need to check up to the square of each number, and you can cache each calculation as you go along.

Problem 21

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

Evaluate the sum of all the amicable numbers under 10000.

def d(n)
  total = 1
  (2..Math.sqrt(n)).each do |i|
    if n % i == 0
      total += n / i + i
    end
  end
  total
end

def amicable?(a, repo)
  repo[a] = b = d(a)
  a != b and a == repo[b]
end

repo  = {}
total = 0

10_000.times do |i|
  if amicable?(i, repo)
    total += (i + repo[i])
  end
end

puts total